Metamath Proof Explorer


Theorem fprodle

Description: If all the terms of two finite products are nonnegative and compare, so do the two products. (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses fprodle.kph k φ
fprodle.a φ A Fin
fprodle.b φ k A B
fprodle.0l3b φ k A 0 B
fprodle.c φ k A C
fprodle.blec φ k A B C
Assertion fprodle φ k A B k A C

Proof

Step Hyp Ref Expression
1 fprodle.kph k φ
2 fprodle.a φ A Fin
3 fprodle.b φ k A B
4 fprodle.0l3b φ k A 0 B
5 fprodle.c φ k A C
6 fprodle.blec φ k A B C
7 1red φ k A B 0 1
8 nfra1 k k A B 0
9 1 8 nfan k φ k A B 0
10 2 adantr φ k A B 0 A Fin
11 5 adantlr φ k A B 0 k A C
12 3 adantlr φ k A B 0 k A B
13 rspa k A B 0 k A B 0
14 13 adantll φ k A B 0 k A B 0
15 11 12 14 redivcld φ k A B 0 k A C B
16 9 10 15 fprodreclf φ k A B 0 k A C B
17 1 2 3 fprodreclf φ k A B
18 17 adantr φ k A B 0 k A B
19 1 2 3 4 fprodge0 φ 0 k A B
20 19 adantr φ k A B 0 0 k A B
21 4 adantlr φ k A B 0 k A 0 B
22 12 21 14 ne0gt0d φ k A B 0 k A 0 < B
23 12 22 elrpd φ k A B 0 k A B +
24 6 adantlr φ k A B 0 k A B C
25 divge1 B + C B C 1 C B
26 23 11 24 25 syl3anc φ k A B 0 k A 1 C B
27 9 10 15 26 fprodge1 φ k A B 0 1 k A C B
28 7 16 18 20 27 lemul2ad φ k A B 0 k A B 1 k A B k A C B
29 3 recnd φ k A B
30 1 2 29 fprodclf φ k A B
31 30 mulid1d φ k A B 1 = k A B
32 31 adantr φ k A B 0 k A B 1 = k A B
33 5 recnd φ k A C
34 33 adantlr φ k A B 0 k A C
35 29 adantlr φ k A B 0 k A B
36 9 10 34 35 14 fproddivf φ k A B 0 k A C B = k A C k A B
37 36 oveq2d φ k A B 0 k A B k A C B = k A B k A C k A B
38 1 2 33 fprodclf φ k A C
39 38 adantr φ k A B 0 k A C
40 30 adantr φ k A B 0 k A B
41 9 10 35 14 fprodn0f φ k A B 0 k A B 0
42 39 40 41 divcan2d φ k A B 0 k A B k A C k A B = k A C
43 37 42 eqtrd φ k A B 0 k A B k A C B = k A C
44 28 32 43 3brtr3d φ k A B 0 k A B k A C
45 nne ¬ B 0 B = 0
46 45 rexbii k A ¬ B 0 k A B = 0
47 rexnal k A ¬ B 0 ¬ k A B 0
48 nfv j B = 0
49 nfcsb1v _ k j / k B
50 49 nfeq1 k j / k B = 0
51 csbeq1a k = j B = j / k B
52 51 eqeq1d k = j B = 0 j / k B = 0
53 48 50 52 cbvrexw k A B = 0 j A j / k B = 0
54 46 47 53 3bitr3i ¬ k A B 0 j A j / k B = 0
55 nfv k j A
56 1 55 50 nf3an k φ j A j / k B = 0
57 2 3ad2ant1 φ j A j / k B = 0 A Fin
58 29 3ad2antl1 φ j A j / k B = 0 k A B
59 simp2 φ j A j / k B = 0 j A
60 52 biimparc j / k B = 0 k = j B = 0
61 60 3ad2antl3 φ j A j / k B = 0 k = j B = 0
62 56 57 58 59 61 fprodeq0g φ j A j / k B = 0 k A B = 0
63 62 rexlimdv3a φ j A j / k B = 0 k A B = 0
64 63 imp φ j A j / k B = 0 k A B = 0
65 0red φ k A 0
66 65 3 5 4 6 letrd φ k A 0 C
67 1 2 5 66 fprodge0 φ 0 k A C
68 67 adantr φ j A j / k B = 0 0 k A C
69 64 68 eqbrtrd φ j A j / k B = 0 k A B k A C
70 54 69 sylan2b φ ¬ k A B 0 k A B k A C
71 44 70 pm2.61dan φ k A B k A C